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A PRACTICAL TREATISE 



YARN ..^^ CLOTH CALCULATIONS 



COTTON FABRICS 



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THOMAS YATES, 

Instructor in Weaving, IVarp Preparation and Calculations 
New Bedford Textile School 



NEW BEDFORD, MASS. 
1901 






n>iiii r »Mi 'm i ) i jp i » ii 



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CLAR« Oi^XYo Mo, 
COPY R. 



Entered according to Act of Congress 
in the year 1901, 

By THOMAS YATES, 

In the office of the Librarian of Congress, 
Washington, D. C. 



2-10^6/ 






PREFACE 



In the composition of this short wort no attempt 
has been made to cover the subject in all its details^ 
but an effort to meet a demand for a useful book of 
reference, and to those engaged in the different 
branches of cotton manufacture this book is respect- 
fully submitted. 

mOMAS YATES. 
New Bedford, 1901 



YARN AND 
CLOTH CALCULATIONS. 



In dealing with Yarn Calculations, it is 
necessary to have two tables or standards, 
one of length and one of weight. 

The standards of length for different 
materials vary, but the calculations are all 
based on a given length that is contained in a 
o'iven weight. 

Taljles of lengtlis and weights will be given 
with rules for calculations under the specified 
materials. 

The table of lengths for Cotton Yarn is as 
follows : 

1 1 yds.= 1 thread or cir. ot leel, 
120^ " = 80 " = I skein or lea 
840 '• =560 " =7 '• '' '' = thank 

The term ^'hank" in the above table does not 
mean that the yarn is in hank form, as it may 
be in any of the usual forms, such as cops, 
bobbins, skeins, hanks, or warps, etc., but is 
used to indicate a definite lengtli of yarn, i. e. 
840 yards. 

Table of weights for Cotton Yarn is as 
follows : 

437^ grains =: 1 oz., 

7000 " = 16 oz. = 1 lb. 



The above pound is the ordinary system, 16 
oz. = 7000 grains, or 1 lb. avoirdupois, and 
is the standard weight used for all materials. 
The table of lengths e:iven is the standard for 
cotton in the United States and England. 

Cotton Yarns are indicated by numbers or 
counts, the latter indicating how many lengths 
of 840 yards, or hanks, weigh 1 lb. ; for as 
many hanks as weigh 1 lb. equal the counts. 

Example. If 1680 yards weigh 1 lb. 
the counts are 2s. 

1680 -^ 840 = 2 -^ 1 = 2s counts. 

Thus it will be seen that the following rule 
may be used : 

To find the counts of a Cotton Yarn 
when the length and weight is known : 

Rule. DivicJe the length hij the Mfwdnrd 840, 
and, divide the result by the iveight in lbs. 

Example. 16800 yards of Cotton Yarn 
weigh 1 lb. Find the counts. 

16800 -^ 840 ^ 1 = 20s counts. 
Example. If 33600 yards of Cotton Yarn 
weigh 5 lbs., what are the counts? 

33600 ^ 840 -:- 5 r= 8s counts. 
Example. 80 bundles of yarn each con- 
tain 2100 yds., weight 20 lbs. in all Find 
the counts ? 



2100 X 80 =168000 yds. 

168000 -^ 840 -f- 20 = 10s counts. 

Example. A warper beam contains 420 
ends, 18000 yds. long, and weighs 600 lbs. 
net. What are the counts ? 

18000 X 420 = 7560000 yds. 
7560000 ^ 840 ^ 600 =^15s counts. 

The preceding examples show that j'arn in 
any form may be reduced to total yards, and 
by dividing by 840 it may be changed to 
hanks : by dividing hanks by weight the counts 
of the yarn are found. 

In the mills, the counts of yarn are found 
by reeling 120 yards from cops, bobbins, etc., 
on a wrap reel, and dividing its weight in 
grains into 1000, the answer being the counts 
of the yarn. 840 yards is too much to reel, 
so one skein or one-seventh of a hank is taken, 
and its weight divided into 1000, which is 
one-seventh of a lb. of 7000 grains. 

Example 120 yards or 1 skein weighs 
20 grains. What are the counts ? 

1000 -^ 20 =r 50s counts. 

If 2 or more cops or bobbins are reeled and 
weighed together, a proportionate part of 
7000 must be used for a dividend ; if 2 cops 
are reeled, 1 skein from each, and weighed 
together, their weight divided into 2000 will 



give the counts ; if 4 cops are taken, then 
4000 must be used for a dividend, etc. 

Example. 4 cops are taken and reeled 120 
yds. from each, and weighed together; the 
weight is found to be 80 grains, thus 
4000 -f- 80 = 50s counts. 
Example. 2 cops are reeled one skein 
from each, and together weigh 50 grains. 
What are the counts ? 

2000 -4- 50 =r 40s counts. 
Any other length or portion of a hank may 
be taken, using the same portion of 7000 for a 
dividend. This will be better understood by 
reference to the following table of lengths and 
their dividends : 

840 yds. = 7 skeins = 1 hank 7000 dividend 
480 " =r 4 '^ = i '^ 4000 '' 
240 ^^ — 2 " = I '' 2000 '' 
120 '^ = 1 '^ = i " 1000 " 
60 " = i '^ = ^L u 500 " 
30 " =1" = oV " 250 " 

15 105 ""•" 

^ —120 —840 ^'^^ 

The number of yards of Cotton Yarn that 
weigh 8i grains equals the counts of the 
yarn. 

To find the dividend for any length of 
yarn; 



Rule. MiiJtijjhj t/te length by 100, a7id 
divide the residt by 12 ; answer is the dividend. 

Example. Find dividend to use when 96 
yards of Cotton Yarn are weighed. 

96 X 100^ 12 = 800 dividend. 
Example. 72 yards of Cotton Yarn weigh 
12 grains. What are the counts ? 

72 X 100 ^ 12 = 600 dividend. 
600 -f- 12 = 50s counts. 
Another rule which will be found useful 
when only short lengths of threads are avail- 
able is as follows : 

Rule. The nwnber of inches of cotton yarn 
that iveigh one grain multiplied by .2314 ecjuals 
the counts. 

Example. 63 threads, each 3 inclies long' 
weigh 1 grain. What are the cotton counts ? 
63 X 3 = 189 inches. 
189 X .2314 = 43.73 counts. 

To find the length in Yards when the 
counts and weight are known : 

Rule. Multijdy the standard length, 840 
yards, by the counts, and the result by the ireight 
in jJowids. 



10 

Examples. 

1. What is the length of yarn in 10 
pounds of 20s Cotton Yarn ? 

840 X 20 X 10 = 168000 yards. Ans. 

2. What is the length of yarn contained 
in 65 pounds of 18s Cotton Yarn? 

840 X 18 X 65 == 982800 yards. Ans. 

3. What is the length of 25 pounds of 40's 
Cotton Yarn ? 

840 X 40 X 25 = 840000 yards. Ans. 

4. If 32 cops of 55s yarn, each cop con- 
taining an equal amount of yarn, weigh 1 
pound, what is the number of yards on each 
cop? 

840 X 55 = 46200 yds. in 1 lb. of 55s yarn. 
462004-32 = 1443.75 yds. on each cop. Ans, 

Example. A spool, togetlier with the yarn 
which it contains, weighs 18 ounces. The 
weight of the empty spool is 8 ounces, the 
counts of the yarn is 28s. What is the length 
of the yarn on the spool ? 
18 — 8 = 10 ounces of yarn. 
840X28 = 23520 yds. in 1 lb. or 7000 grains. 
23520 ^ 7000 = 3.36 yards in 1 grain. 

There are 437.50 grains in 1 ounce. 
437.50 X 10 ozs. = 4375 ozs. weight of yarn. 
4375 X 3.36 yds. in 1 gr. = 14700 yds. on spool 



11 

Example. A warper beam contains 420 
ends of 15s yarn, and weighs 600 lbs. net. 
What is the length of the yarn on the beam ? 
840 X 15 X 600 = 7560000 yds. continuous 

thread. 
7560000 ^ 420 = 18000 yards. 

To find the weight when the length and 
counts are known; 

Rule. Divide tite lengtli of yam in yanh by 

the standard 840, and divide the result by the 

counts. 

Examples. 

1. What is the weight of 168000 yards of 
number 20s cotton yarn ? 

168000 "-f- 840 = 200 
200 ^ 20 = 10 pounds. Ans. 

2. What is the weight of 18480 yards of 
44s cotton yarn ? 

18480 -^ 840 = 22 
22 -i- 44 = .5 or i pound. Ans. 

3. What is the weight of 302400 yards of 
36s cotion yarn ? 

302400 -^ 840 = 360 
360 -^ 36 :=: 10 pounds. Aus. 

4. A warper beam contains 500 ends of 
32s yarn and is 24000 yards long. What is 
the weight of the yarn ? 



12 

24000 X 500 = 12000000 yards. 
12000000 ^ 840 = 14273.81 hanks. 
14273.81 -^ 32s = 446.05 pounds. Ans. 
Cotton in process of manufacture is worked 
througli the pickers, cards, and drawing frames, 
without having twist put nito it, but from the 
slubber to the spinning frame or mule, the 
amount of twist put in increases with every 
process. 

Twist is put in the roving and yarn to give 
strength, and is calculated by the number of 
turns put in one inch. 

The turns per inch vary according to the 
quality of cotton u?ed, and the class of yarns 
being made, and is determined by multiplying 
the square root of the counts by a constant. 
The constant for warp yarn in the U. S. is 4.75 
" '' ^' filling yarn ^' '' '' 3.25 

^' •' '' hosiery yarn '• " " 2.50 

To find the turns of twist per inch : 

Rule. MuUiphj the square root of the counts 
hy the co7istant. 

Example. Find the turns per inch on a 25s 
Avarp yarn using 4.75 for a constant. 

y'25 = 5 the square root of 25 is 5. 
5 X 4.75 = 23.75 turns per inch. 

Warp yarns are tested for strength by 
breaking one skein on a testino; machine. 



13 

To find the standard breaking weight of 
warp yarn : 

Rule. Divide 1760 hy the counts of yarn ; 
answer is the standard hreakinor weis'ht in lbs. 

Example. What is the standard breaking 
weight of a 40s warp yarn ? 

1760 -f- 40 = 44 lbs. 

Although 1760 is accepted as the standard 
constant for breaking weight, it will be found 
in cloth mills, where fine yarns are woven, that 
a higher constant must be used to get satis- 
factory results in the weave room ; while in 
mills on coarse counts a lower constant may 
be used and satisfactory results follow. 



PLY YARNS. 

Ply threads are made by twisting two or 
more threads together; the resulting thread 
being the counts of the threads used divided 
by the number of threads twisted, when the 
single yarns are equal. Thus if two threads 
of 40s were twisted, the resulting thread 
would be equal to a 20s, and would be in- 
dicated as a 2-40s. If three threads of 45s 
were twisted, it would be indicated as a 3-45s 
and would be equal to a 15s thread. 



14 

In fact the resulting- counts are a trifle 
coarser than the indicated counts, as the yarn 
contracts in the twisting about 5 per cent. 

RESULTANT COUNTS. 

To find the counts when threads of dif- 
ferent counts are twisted together- 
Rule. Multiphj the two counts together, and 
divide by their sum. 

Example. If a thread of 30s and one of 
60s are twisted together, what are the result- 
ant counts ? 

30 X 60 = 1800 ^^ 

= 20s counts. 

30 + 60 = 90 

To find the counts of a ply thread, when 

three or more threads of different counts 

are twisted together- 
Rule. Divide the highest counts by its own 

cou7its and by each oj the others ; add the residts, 

and divide into the highest counts. 

Example. A 20s, 40s and 60s cotton 

thread are twisted together. What are the 

resultinijf counts ? 



60 
60 
60 



60 = 1 
40 1= 1.5 
20 = 3 



5.5 
60 -h 5.5 = 10.9s counts of ply thread. 



15 

To find the counts of a yarn which must 
be twisted with another to make a ply 
yarn of a given count- 
Rule. Multiijly the covnts of the desired ply 
yar7i by the cotmts of the known single yarn, and 
divide the restdt by the difference in the two 
counts. 

Example. What are the counts of a 
thread that must be twisted with a 44s cotton 
thread to produce a ply thread equal to a 24s 
single ? 

44 X 24 = 1056 .^ . , . , 

= 52.8s counts required 

44 _ 24 = 20 ^ 

To find the weight of each counts of yarn 
when twisting two threads together to ob- 
tain a given weight of ply yarn- 
Rule. First find, the counts of the ply yar7i. 
The7i multiply the counts of the ply yarn by the 
total iveight required, arid divide the residt by 
the counts of either of the single yarns. The 
answer will be the re(piired iveight of the single 
yarn used for a divisor. 

Example. What weight of 40s cotton 
yarn would be twisted with a 32s to produce 
100 lbs. of ply yarn ? 
40X32 = 1280 .„^^ . f.i. 1 

40+32 ^-72"^ -^^'^^^^^^"^^^^^^^P^^^^^^ 



16 

Then : 

17.77X100 = 1777 
1777 ^40s = 44.44 lbs. of 40s yarn 
100 —44.44 = 55.56 " " 32s " 
The rule applies equally well when more 
than two threads are twisted. 

Example. What weight of each of 120s, 
80s and 40s would be used in making up 100 
lbs. of ply yarn ? 

First find the counts of the ply yarn, as 
follows ; 

120 -^ 120 .= 1 
120 -^ 80 = 1.5 
120 ^ 40 = 3 

120 -f- 5.5 = 21.82s counts of the ply yarn. 

Then : 
21.82X100 r= 2182 
2182 -^120s = 18.181bs.wgt. of 120s yarn. 

Then : 
2182 -f- 80s = 27.27 lbs. wgt. of 80s yarn. 

Then : 
2182 -:- 40s = 54.55 lbs. wgt. of 40s yarn. 

To find the cost per lb. of a ply yarn 
made from two or more single yarns- 
Rule. Find the weight of each of the single 
yarns used, (by the last rule giveii) and multiply 



17 

the weigJtt of each yarn bij its own cost; add the 
results, and divide hij the number of lbs. of ply 
yarn made ; the answer is the cost per lb. of the 
ply yarn. 

SPUN SILK. 

Spun Silk is calculated by the same method 
as Cotton Yarn, the same tables of lengths 
and weights being used. Thus the rules given 
for calculating single cotton yarns will apply 
to spun silk. Ply yarns, however, are indica- 
ted in a different manner. 

Cotton ply yarns are indicated by placing 
the number of threads twisted before the num- 
bers of the single yarn ; a two ply 40s would 
be indicated 2-40s, and would be equal to a 
single 20s. In spun silk, the method is to 
place the actual size or counts of the ply 
thread first, and the number of threads used 
after it. Thus a two ply thread of spun silk, 
which would equal a 40s, would be composed 
of two threads of 80s, and would be indicated 
40s-2 ; a 30s-3 would be composed of three 
threads of 90s. 

Thus the count of a ply thread in spun silk, 
multiplied by the number of threads twisted, 
equals the counts of the single threads. 



18 

Example. What counts of single yarn 
would be used to make a 30s-3 spun silk ? 
30 X 3 r= 90s counts. 
Example. What counts of spun silk are 
equal to a 60s cotton ? 

60s counts. 
Example. How many yards are contained 
in 10 lbs. of 20s-2 spun silk? 

840 X 20 X 10 = 168000 yards. 
Example. How many yards of single yarn 
are required to make 10 lbs. of 20s-2 spun 
silk ? 

20s-2 = 40s-l 
Then: 
4:0x840X 10 = 336000 yards of single 40s. 

RAW SILK. 

Raw silk is calculated on an entirely dif- 
ferent basis from spun silk. The American 
custom of specifying the size is to take the 
weight of 1000 yards in one dram avoirdupois. 

Thus if 1000 yards weighed 1 dram, it 
would be 1 dram silk, and would contain 
256000 yards in one pound, because if one 
dram contained 1000 yards — there being 256 
drams in one pound — a pound would contain : 
256 X 1000 = 256000 yards. 



19 

The following table may be used for cal- 
culating raw silk : 

16 drams = 1 oz. = 437.5 grains. 
256 '' = 16 " = 7000 " = 1 lb. 

Raw silk is sometimes indicated by the num- 
ber of yards contained in one pound ; a silk 
that contained 102400 yards in one pound 
would be known as 102400 yard silk, which 
would also equal a 2i dram silk. 

To find the number of idram silk from a 

given length and weight- 
Rule. Multiply 256 hy 1000 and the 

weight in poundsj and divide the result hy the 

total length of yarn. 

Example. 128000 yards of raw silk weigh 

one pound. What is the number of dram silk ? 
256 X 1000 = 256000. 
256000 -^ 128000 == 2 dram silk. Ans: 
Example. 132740 yards of raw silk weigh 

3i pounds. Find the numbers in dram silk. 
256 X 1000 X 3i = 896,000. 

896000 -^ 132740 = 6.75 dram silk. Ans. 

To find the number of yards when the 
weight and counts are known. 

Rule. Multijjly 256 % 1000, and the result 
by the weight in ponnds, and divide hy the counts 
of the yarn. 



20 

Example. Find the number of yards con- 
tained in 5 pounds of 4 dram silk. 
256 X 1000 = 256000 X 5 = 1280000 
1280000 -^ 4 = 320000 yards. Ans. 
Example. Find the number of yards con- 
tained in one pound of one dram silk. 
256 X 1000 = 256000. 
256000 X 1 = 256000. 
256000 -f- 1 = 256000 yards. Ans. 

WORSTED YARNS. 

The table of lengths for worsted yarn is as 
follows : 

1 yard = 1 thread or cir. of worsted reel 
80 yards = 80 " =: 1 lea or knot 
560 yards = 560 ^' = 1 " ^' = 1 hank 

The table of weights is the same as for cotton 
yarn. 

The number of hanks of 560 yards contained 
in one pound, indicates the counts. 

The rules given for cotton yarns apply 
equally well for worsted, except those given 
for finding counts from short lengths. 

To find the dividends for short lengths 

of worsted yarns- 
Rule. Multiply the length hy 100, and 

divide the result hy 8 ; answer is the dividend. 



21 

Example. Find the dividend to use when 
72 yards of worsted yarn is weighed. 
72 X 100 = 7200 
7200 ^ 8 = 900 dividend. 
Example. 56 yards of worsted yarn Aveigh 
30 grains. What are the counts? 
56 X 100 =.5600 
5600 ^ 8 = 700 dividend 
700 -^ 30 = 23.33 counts. 
Example. A beam containing 1568 ends, 
and 300 yards long, weighs 30 lbs. What are 
the worsted counts ? 

1568 X 300 = 470400 
470400 -^ 560 ^ 30 = 28s counts. 

To find counts, length, or weight, apply 
rules given for cotton yarns, and use the 
standard for worsted, i. e. 560 yards. 

WOOLEN YARNS. 

There are two standards for length used in 
the calculations for woolen yarns, viz., the run 
and the cut systems. 

In the cut system 300 yards are taken as 
the standard for length, and the number of 
cuts of 300 yards contained in one pound, in- 
dicates tlie counts. 

In the run system, 1600 yards are taken as 
the standard length, and the number of runs of 



22 

1600 yards contained in one pound, indicates 
the counts. 

The full table of woolen lengths is as 
follows : 

1 yard =1 thread or cir. of the woolen reel 
80 yards =1 knot. 
300 yards =3| knots = 1 cut. 
1600 yards = 20 knots = 5^ cuts = 1 run. 

To find counts, length, or weight, apply 
rules given for cotton and worsted yarns, 
and use the standard length of the system 
desired— 

Example. What are the cut counts of a 
woolen yarn when 1200 yards weigh 2 lbs? 
1200 -^ 300 -f- 2 = 2s cut counts. 
Example. What is the length of 100 lbs. 
of 3 cut yarn ? 

300 X 3 X 100 = 90000 yards. 

Example. Find the weight of 60000 yards 
of 4 cut yarn. 

60000 -^ 300 = 200. 
200 -^ 4 = 50 lbs. Ans. 

Example. What are the run counts ot a 
woolen yarn when 100000 yards weigh 7^ lbs ? 
100000 ^ 1600 = 62.5. 

62.5 -i- 7.5 = 8.33 run yarn. 



23 

Example. What is the length of 75 lbs. of 
2 run yarn ? 

1600 X 2 X 75 = 240000. 
Example. What is the weight of 38400 
yards of a 3 run yarn ? 

38400 ^ 1600 -f- 3 = 8 lbs. 

EQUIVALENT COUNTS. 

When dealing with the numbering of different 
yarns, it frequently becomes necessary to know 
what would be the counts of a certain yarn it 
numbered according to a different standard. 

This is known as converting the counts of 
a thread in one system into equivalent counts 
of another system. 

To find the equivalent counts of a yarn 

in any system- 
Rule. MuUiply the given counts by its own 

standard length, and divide the result by the 

standard yards in the system required. 

Example. What are the equivalent counts 

of a 15s cotton in woolen run counts ? 
840 X 15 = 12600 yards. 
12600 ^ 1600 r= 7.87 counts woolen run. 
Example. Find the equivalent of a 50s 

worsted in cotton counts. 

50 X 560 = 28000 yards. 

28000 -^ 840 = 33.33s cotton counts. 



24 

PRICES OF TWISTED YARNS. 

To find the price of a ply yarn when the 
threads twisted together are of different 
values, and different counts — 

Rule. Multiply the highest covnts hy the 
j)rice of the lowest counts, and the lowest counts 
hy the price of the highest; divide the sum of 
the products hy the sum of the counts. 

Example. A 32s yam costs 29c. per 
pound,- aDd a 16s yarn costs 18c. per pound. 
Find the cost of the two twisted together. 

32 X 18 = 576 

16 X 29 = 464 

48~ T040 

1040 ^ 48~="21|c. per lb. Ans. 

If the threads are of different materials, 
and the counts reckoned on a different basis, 
reduce them to the same denomination, and 
proceed as in the previous example. 

Example. Find the price per pound of a 
twisted thread composed of 1 thread of 40s 
worsted at 75c. per lb. and 1 thread of 60s-t^ 
spun silk at $2.30 per pound. 

First reduce the worsted to spun silk counts, 
40 X 560 -^ 840 = 26.66 spun silk counts, 
then apply previous rule. 



25 

26.66 X $2.30 r= $61,318 
60 X $0.75 =r $45.00 



86.66 106.318 

$106,318 -f- 86.66 = $1.2268 price per lb. 

If three or more threads are twisted together, 
first find the value of any two, and then the 
value of the thread resulting from those two 
with the third. 

Example. Find the price of a three ply 
thread composed of one thread of 40s worsted 
at 80c. per pound, one thread of 28s cotton 
at 30c. per pound, and one thread of a 20 run 
woolen yarn at 56c. per pound. 

First, reduce the worsted to cotton counts, 
and find the price of tlie worsted and cotton 
twisted. 

40x 560 -^ 840 = 26.66 cotton counts. 

Then : 

26 66 X 30c. = $ 7.998 
28 X 80c. = 22.40 

54.66 $30-398 

$30,398 ^ 54.66 = 55 j%\c. price of the 

worsted and cotton thread. 

Next find the counts of the two threads 

twisted by rule previouslv given. 
26.66 X 28 = 746.48 

rrr 13.67 COUUtS. 

26.66 + 28 = 54.60 



26 

Thus we have a yarn equivalent to a 13.67s 
cotton counts costing SSy^J^c. per pound. 

Next reduce the 20 run woolen to cotton 
counts. 

20 X 1600 -4- 840 = 38.10s in cotton counts. 
Thus we have yarns as follows : 
38.10s at 56c. per pound, 
13.67s at 55^^^c. per pound, 
then proceed as above. 

38.10s X 553-VoC. = 121.18 
13.67s X 56c. = 7.65 

51.77 $28.83 

$28.83 -^ 51.77 = 553-yoC. price of 3 ply 
thread. 

CLOTH CALCULATIONS. 

In dealing with cloth calculations, it may be 
best to explain the necessity for the rules 
which are given later. 

As is well known, cloth is made by inter- 
weaving threads that are drawn through loom 
harnesses and reed ; the harnesses raising and 
lowering alternate threads (on plain cloth), 
and the shuttle carrying the filling thread 
through the space called the shed, and is beat 
up by the reed. 

The quality of cloth is determined by the 
fineness of the threads, and the number that 
are contained in one inch. 



27 

As the threads cross each other, they are 
bent more or less out of a straight line, a fact 
which causes cloth to contract both in width 
and length, and the rate per cent of contrac- 
tion will vary on different cloth constructions. 

For this reason, reeds must be used of a 
lower count than the required count of the 
cloth, and this makes the cloth wider at the 
reed than the width of the cloth on the roller. 

The length of a cut must also be made 
longer on the slasher than the required length 
of the cut when woven. 

The threads of the warp must be sized be- 
fore they can be woven, and this adds to the 
weight so that size and contraction are factors 
to be considered in all cloth calculations. 

The rules and examples given in the first 
part of the book refer to plain cloth, twills, or 
sateen, or such cloth that have an equal num- 
ber of threads drawn in each dent of the reed. 
Rules that apply to fancy cloths will follow. 

The principal particulars of a plain cloth 
are the number of threads of warp and filling 
in one inch, and the width of the cloth when 
woven, and the weight, which is designated as 
so many yards per pound. 

The number of warp threads per inch is 
known as the sley, and the filling threads as 



.28 

the picks, and when speakinj^ of a cloth con- 
struction, the sley is invariably put first. A 
cloth tliat contains 56 warp threads and 60 
filling threads per inch, would be classed as a 
56 X 60 cloth. Warp threads are commonly 
known as ends, and this term will be used in 
all the calculations. 

To find the number of ends required in 
a warp when the sley and width are 
given — 

Rule. Multiply the sJcy hij the ividth and 
add extra ends for the selvages. 

Selvages are made by drawing some of the 
ends on the sides double, the number so 
doubled will depend on the width of selvage 
desired ; extra ends are added to keep the 
cloth at the desired width. On heavy cloths, 
sometimes double twisted ends are used for 
selvages. 

Example. How many ends are required to 
weave a cloth 56 sley and 36 inches wide with 
40 ends extra for selvage ? 

56 X 36 =2016 + 40 = 2056 ends. 

Example. How many ends will be neces- 
sary to weave a cloth 76 sley, 30 inches wide, 
allowing 36 ends extra for selvages? 

76 X 30 r=2280 + 36 = 2316 ends. 



29 

On plain cloths of ordinary construction, it 
is common to allow from 6 to 7 per cent for 
contraction in width, and, as low sley cloths 
contract more than high sley cloths, the follow- 
ing rule is given to find dents per inch in reed, 
by which a sliding rate of contraction is 
allowed. 

To find the dents per inch in a reed re- 
quired for any sley cloth. 

Rule. Subtract 1 fiom the sley, then subtract 
5 per cent, divide the result by the number of 
ends drawn in 1 dent, and the answer is the dents 
per inch required. 

Example. Find the dents per inch in a 
reed, to weave a 48 sley cloth, ends drawn 2 
in a dent. 

48 — 1 = 47 — 5 per cent = 44.65 
44.65 -^2 = 22.32 dents per inch. 
If no contraction took place, a reed for a 
48 sley cloth would require 24 dents per inch ; 
the rate per cent allowed by the above rule is 
found to be 7 per cent. 

24 — 22.32 = 1.68 dents less. 
1.68 X 100 ^ 24 = 7 per cent. 
Example. Find the dents per inch in a reed, 
to weave a 96 sley cloth, ends drawn 2 in a 
dent. 



30 

96 — 1 = 95 — 5 per cent = 90.25 
90.25 -f- 2 = 45.12 dents per inch which 
gives 6 per cent for contraction. 

48 — 45.12 = 2.88 dents less. 
2.88 X 100 -^ 48 = 6 per cent. 
Example. Find the dents per inch in reed, 
to weave a 112 sley cloth, with ends drawn 3 
in a dent. 

112 — 1 = 111 — 5 per cent = 105.45 
105.45 -^ 3 = 35.15 dents per inch. 

To find the sley of the cloth a reed will 
weave when length and total dents in reed 
are known. 

Rule. Divide the total dents by the length of the 
reed; the answer is the dents in 1 inch; then mul- 
tiplij by the number of ends to be drawn in a dent, 
add 5 per cent to the result, then add 1 and 
answer is sley of cloth. 

Example. Find the sley cloth a reed will 
weave that contains 1080 dents in 36 inches, 
with ends drawn 2 in a dent. 

1080 -^ 36 = 30 dents in 1 inch. 
30 X 2 = 60 

60 + 5% = 63 + 1 = 64 sley cloth. 
Example. What sley cloth will a reed weave 
that contains 23.12 dents in 1 inch, the ends 
to be drawn 3 in a dent. 



31 

23.12 X 3 = 69.36 
69.36 + 5% = 72.83 + 1 = 73.83 or 74 sley. 
When making calculations to find the amount 
of filling in a given length of cloth, the length 
of the filling pick must be taken at the reed, 
and not the actual width of the cloth. This 
makes it necessary to find width at reed. 

To find width of the warp ends in the 
reed. 

Rule. Divide the total ends in the warp by 
the number of ends in a dent, subtract from the 
result as many dents as contain double ends at 
the selvages, and divide the result by dents per 
inch in the reed. 

Example. Find the width in the reed on a 
64 sley cloth, 36 inches wide, drawn 2 in a 
dent, with 48 extra ends drawn in 12 dents for 
selvages. 

64 X 36 + 48 3= 2352 ends 
2352 -f- 2 — 12 = 1164 dents that con- 
tains ends. 

64— 1 = 63 — 5% = 59.85 

59.85 -^ 2 =: 29.92 dents per inch in reed. 

1164 ^ 29.92 = 38.90 inches at reed. 

The contraction which takes place in the 
length of the warp is caused by its being bent 
out of a straight line by the filling. 



32 

The coarser the filling, and the more picks 
put in the cloth, the greater will be the con- 
traction from length of warp to cloth. The 
following rule will therefore explain itself: 

To find the slasher length of a cut for a 
given length of cloth- - 

Rule. Midtiphj the picks per inch by 4, and 
divide the result by the counts of the fillings the 
answer is the per cent to allow for contraction. 

Example. What is the slasher length of a 
cut that is woven with 68 picks per inch with 
32s filling, the cut to be 50 yards long when 
woven. 

68 X 'i ^- 32 = 8.5 per cent, 

50 X 8.5 per cent = 4.25 yards, 

50 + 4.25 ^ 54.25 yards slasher length. 

To find the weight of warp yarn in a 

piece of cloth- 
Rule. Multiply the numher of ends in the 

piece by the slasher length, divide the result by 

the standard yards and the counts of the yarn ) 

answer will be weight of yarn without size. 
If the weight is desired when sized add 7 

per cent. 

Example. A piece of cotton cloth 68 sley, 

36 inches wide, contains, with selvage ends, 

2488 ends of 36s yarn. 



33 

What is the weight of warp yarn contained 
in 50 yards of cloth, with 7 per cent of size 
and 5 per cent for contraction ? 

50 -f 5% for contraction = 52.50 yards 
slasher length. 

2488 X 52.50 = 130620 total yards, 

130620 -^ 840 -^ 36s = 4.31 lbs. 

4.31 +7% size= 4.61 lbs. total wgt. 

To find the weight of filling— 

To find the weight of filling in a given 
length of cloth, the width of the warp at the 
reed must be taken as the length of the filling 
pick, and not the actual width of the cloth. 

Rule. Multlphj tlie width at the reed by 
picks j)er inch, and the result by the length of 
jiiece in yards, and divide by the standard yards 
and the counts. 

Example. A cut of cotton cloth 50 yards 
long is woven with 72 picks per inch, with 60s 
filling and is 32.50 inches wide at the reed. 
Find the weight of the filling. 
32.50 X 72 = 2340 inches of filling in 1 
inch of cloth ; it also equals the number ot 
yards of filling in 1 yard of cloth. 

Then : 
2340 X 50 -^ 840 -^ 60 = 2.32 lbs. of filling. 



34 

Example. A cotton cloth constructed 
64 X 68, and 36 inches wide, 45s filling, 48 ends 
extra for selvages drawn in 12 dents. 

Find weight of filling in 50 yards of cloth. 
64— 1 = 63 — 5%= 59.85. 
59.85 -^ 2= 29.92 dents per inch in reed. 
64 X 36 + 48 = 2352 ends. 
2352 -^ 2 — 12 = 1176 dents used. 
1176 -^ 29.92 = 39.30 width at reed. 
39.30 X 68 X 50 = 133620 yards of filling. 
133620 ^ 840 -^ 45 = 3.53 lbs. ot filling. 



AVERAGE NUMBERS. 

One of the principal factors in cloth calcu- 
lations is the weight, which is usually designa- 
ted in the cotton trade as so many yards per 
lb., and ^^'hen the sley, pick and width of a 
cloth and yards per lb. are known, the required 
numbers of yarn to make it, are first deter- 
mined by findnig the average number i. e. a 
number or counts for warp and filling that 
would give the weight desired. 

On the coarser grades of cloth, the differ- 
ence in the counts of warp and filling is slight, 
but as the cloth becomes of a finer grade the 
difference between the counts of warp and 
filling is greater. 



35 

To find average number when sley, pick, 
width, and yards per lb. are known. 

Rule. Multiply number of ends, by slasher 
length, divide result by standard yards, (840 
for cotto7i) aiiswer is hanks of warp yarn. Add 
7 per ce7it for size, and answer represents hanks 
of warp yarn with size added. Multiply width 
at reed by picks per inch, and result by length oj 
cut, divide answer by standard, yards, (840 for 
cotton) answer is hanks of filling. 

Divide length of cut by yards per lb., answer 
is weight of cut. 

Add hanks of luarp and filling, and divide by 
weight of cut, ansiver is average numbers. 

Example. Find average numbers on a cloth 
48 X 52, and 36 inches wide, to weigh 6 yards 
per lb., and 50 yards to cut. 

48 X 36 +40 = 1768 ends. 

1768 X 54 yds., slasher length, = 95472 
yds. warp. 

95472 ~ 840 + 1% for size = 122.74 
hanks warp. 

39 inches at reed X 52 X 50 = 101400 
yards filling. 

101400 -^ 840 = 120.71 hanks filling. 

122.74 + 120.71 = 243.45 total hanks. 

50 yds. of cloth ~ 6 yds. per lb. = 8.33 
wgt. of cut. 



36 

243.45 -4- 8.33 = 29.22 avg. numbers. 
Another i-ule is given for finding average 
numbers wliich is much shorter. It is also 
correct and practical. 

To find average number when sley, pick, 
width, and yds. per lb. are known- 
Rule. Add the sley and jricJc, and multiphj 
the result by the width, and yards per lb. 
Divide the result by 750 on ivarj) counts below 
50s, and by 760, when above those counts. 

The explanation of the above rule is as 
follows : 

Sley + pick X width X yards per lb. equal 
the number of yards of yarn in 1 lb. ol 
cloth. The standard for cotton yarn is 840 
yards, but, as the yarn contracts in weaving, it 
adds weight, and the sizing on the warp yarn 
also adds weight. This increased weight on 
coarse yarns equals about 10 per cent. This 
percentage, added to the standard yards, makes 
them heavier, and a less number of yards will 
weigh 1 lb., and thus, instead of using 840 for 
the standard, less yards, i. e., 750 yards are 
used. Fine yarns contract less and also carry 
less size than coarse yarns, therefore 760 is 
used for a constant. 



37 

Example. Find the average numbers on a 
cloth 52 X 56 and 36 inches wide to weigh 4 
yds. per lb. 
52 + 56 X 36 X 4 ^ 750= 20.73 avg. No. 

To find the average number when sley, 
pick, warp and filling are known- 
Rule. Divide tJte slnj bij the counts of the 
ivarp, and tJie 'piclis hij the filling ; add the re- 
sults, and divide into the sum of the pich and 
sley. 

Example. A cloth is woven 68x72 with 
with 45s warp and 60s fillinii: : 
Wliat are the avera^'e numbers? 
68 -f- 45 = 1.51 
72 ^ 60 = 1.2 
1.51 + 1.2 = 2.71 

68 + 72 = 140 -^ 2.71 = 51.66 aver- 
age numbers. 

To find average number of warp, when 
two or more different counts are used- 
Rule. Divide the nvmber of ends of each 
kind by its own counts, add the results, and divide 
into the total number of ends. 

Example. A piece of cloth is woven 32 
inches wide, with 84 picks per inch of 75s 
filling, and contains 3200 ends of 80s, 400 
ends of 40s, and 300 ends of 2-30s. 



38 

What is the average numbers ? 

First find average warp. 

3200 -^ 80 = 40 

400 -^ 40 = 10 

300 ^ 2-30 or 1-15 = 20 

70 
3200 + 400 + 300 = 3900 ends 
3900 -f- 70 = 55.71 average warp. 
Then: 

3900 -f- 32 = 122 average sley, and we 
have 122 X 84 

55.71 Warp 
75 Filling. 
122 -i- 55.71 = 2.19 
84-^ 75 = 1.12 
2.19+1.12 =3.31 
122 +84 = 206 
206 -i- 3.31 =: 62.23 average number. 

To find yards per lb., when sley, pick, 
counts of warp, filling, and width are 
known- 
Rule. First Jind average number by method 
jirevioKshj given, then, multiply ilie average num- 
ber by 750 (^if average No. is below 50s) and 
by 760 [if above), divide the result by the sum 
of the sley and pick, multiplied by the width of 
cloth. 



39 

Example. A cloth is constructed 48 x 52 
with 18s warp and 22s filling, 36 inches wide. 
Find the yards per lb. 

48 -^ 18 = 2.66 
52 ^ 22 = 2.36 
2.66 + 2.36 = 5.02 
48 + 52 r= 100 -^ 5.02 = 19.92 avg. No. 

Now apply the second part of rule : 

19.92 X 750 = 14940 

48 + 52 X 36 = 3600 

14940 ^ 3600 = 4.15 yards per lb. 

To find the yards per lb. from a small 
piece of cloth- 
Rule. Mnlliphj 7000 grains hy the mnnher 
of square inches of cloth weighed. Divide the 
resnlt hy the iveight of the piece iveighed in, 
grains, multiplied by 36 and the icidth of cloth 
desired. 

Example. A piece of cloth 3 inches square 
weighs 11 grains; if cloth is made 32 inches 
wide, what are the yards per lb ? 

3 X 3 = 9 sq. in. X 7000 = 63000 
32 X 36 X 11 = 12672 
63000 -^ 12672 = 4.97 yds. per lb. 
More accurate results are obtained by weigh- 
ing larger pieces, and, when possible, 1 yard is 
taken, and its weight in grains divided into 



40 

7000 ; the result is yards per lb. ; or any part 
of a yard may be taken, and its weight in grains 
divided into an equal portion of 7000 for a 
dividend. 

Example. A yard of cloth weighs 1600 
grains, what are the yards per lb ? 

7000 -^ 1600 = 4.37 yds. per lb. 

To find filling required to govern yards 
per lb.— 

In mills where a large variety of cloth is 
made, it is the practice to spin only 3 or 4 
different counts of warp yarn, and regulate 
the weights of the cloth by varying the counts 
in the filling. 

It is economy to manufacture by this 
method, on account of the various processes 
w^arp yarn has to pass tlirough, while filling is 
taken direct from the spinning frame or mule. 

To find filling required to govern yards 
per lb. when sley, picks, width and yards 
per lb. are known- 
Rule. First Jin d the average numbers, then 
divide the sum of sley and picks by average num- 
bers, and answer is iceight of yarn in average 
counts, and represents total loeight ; then divide 
the sley by the counts of the warp, and answer is 



41 

the 'portion of weight that is ivarj) ; deduct 
weight of warjJ from total iveight, and remainder 
is tJie portion ivhich is filling; divide the picks hy 
this jiortion, and answer is counts of filling re- 
quired. 

Example. A cloth is woven 72 x 76 with 70s 
warp and 36 in. wide, weighs 11 yards per lb. 
What counts of filling are required? 

Note. As counts of warp are above 50s 
use 760 for constant. 

72 + 76x36 Xll -^760 =77.11 Avg. No. 

72 -f 76 -^ 77.11 = 1.92 total weight. 

72 ^ 70 = 1.03 weight of warp. 

1.92 — 1.03 = 0.89 weight of lillino;. 

76 -1- 0.89 r= 85.39 counts of lilling re- 
quired. 

On many fancy cloths the ends are not 
drawn in the reed in regular order, and some 
dents contain more ends than others ; therefore 
other methods than those previously explained, 
must be used to find the number of ends in a 
warp, and the counts of the reed to use. It 
may be stated that in such cases all calcula- 
tions are based on the reed draft. 

To find the number of ends in a warp 
for a fancy cloth of irregular reed draft- 
Rule. Multiply the number of ends in one 
pattern by the number of patterns in the full 



42 

width oj the doth ; add extra ends for selvages, 
and result is numher of ends required ; if any 
fraction of a pattern is used add the ends in. 

To find the number of patterns when 
the sley reed, and width of the cloth are 
known— 

The term sley reed is understood to mean a 
reed that will weave a given sley cloth with 
the ends drawn 2 ends in a dent. 

Rule. Divide the sley of the reed by 2, 
midtiply the result by ividth of cloth desired; 
divide this result by the number of dents contain- 
ed in 1 pattern. 

Example. If the following reed draft were 
drawn on a 68 sley reed, how many ends would 
be required in the warp to weave a cloth 30 
inches wide ? 

Dents. Ends. 

16 32 

4 16 

2 4 

4 16 

26 68 

68 -^ 2 X 30 r=r 1020 dents yarn will occupy. 
1020 -i- 26 = 39 patterns and 6 dents over. 



43 

These 6 dents may be utilized by adding 
them to the first 16 dents in the pattern, and 
using 1 1 of them the first time the pattern is 
drawn, which will leave 11 dents to draw at 
the finish of the 39 patterns, thus making both 
sides of the cloth next the selvages alike ; 

Then : 

68 ends in 1 pattern X 39 patterns = 2472 
-|-12 ends for the 6 dents over +48 ends 
for selvages = 2534 total ends. 

Finding the reed to use when designing new 
patterns, or when reproducing a piece of 
woven cloth is not a difficult matter when some 
part of the pattern consists of a plain weave, 
drawn 2 ends in a dent. When no part of 
the pattern is drawn plain, good judgment is 
required to determine how many ends occupy 
a dent, and how many dents occupy a given 
space. 

A series of reed tables have been carefully 
prepared, showing the number of dents that 
are contained in ^L of an inch, as represented • 
in the cloth, and advancing by 16ths up to one 
inch ; the actual dents per inch in the reed is 
also given. 

To illustrate the utility of these reed tables 
take the last example given, and suppose the 
26 dents and 68 ends in 1 pattern measure }| 



44 

of an inch. By reference to the tables, it will 
be found that a 68 sley reed contains 25.50 
dents on that space, but as fractions of a dent 
cannot be used, a 68 sley is the nearest reed 
that can be used. 

When a pattern is so small that one pattern 
vv^ould be too small to measure, it is better to 
count the number of patterns that are con- 
tained in 1 inch. Such patterns as Bedford 
cords come under this head, and the finding of 
a reed for a Bedford cord pattern will be il- 
lustrated by an example. 

Example. Suppose it has been decided to 
draw the ends in the reed as follows : 



Dents. 


Single Ends 


Ply Ends. 


1 


3 


1 


1 


2 


1 


1 


2 


1 


1 


3 


1 


4 


10 


4 


11 pat 


terns measure 1 


inch find the sley 



reed to use. 

4 X 11 = 44 dents in 1 inch. 

Referring to the reed tables, it will be found 
that an 88 sley reed contains 44 dents on 1 
inch, so 88 sley reed is the reed to use. 

On irregular reed drafted cloths, the sley 
of the cloth is spoken of as the average sley. 



45 

To find the average sley on irregular 

drafted cloths- 
Rule. Divide the total ends hy the width of 

the dot] I. 

Sometimes it is desired to change the aver- 
age sley of a pattern without rearranging the 
draft, and in this case, 

To find the sley reed to use to make a 
cloth any average sley desired- 
Rule. Multiphj the average sley required by 
the dents in 1 i)attern, divide the result by the 
number of ends in one imttern ; muUijdy the re- 
sult by 2. Answer is sley reed to use. 

Example. Find the sley reed required to 
make a cloth average 110 sley with the follow- 
ing reed draft. 

Dents. Ends. 

6 12 

1 4 

1 2 

1 4 

9 22 

110 X 9 ^ 22 = 45 X 2 = 90 sley reed. 
Proof. Make the above pattern 30 inches 
wide and find average sley. 



46 

90 ^ 2 X 30 -^ 9 = 150 patterns. 

150 X 22 = 3300 ends. 

3300 -f- 30 = 110 average sley. 

On fancy weaves in which there are 2 or 
more kinds of warp yarn, the weight of each 
must be calculated separately. 

In the case where ply yarns are used, they 
will not be sized, and the rate of contraction 
will be different from the single yarn. 

When cords formed by drawing several 
single ends in 1 harness eye, or when ply 
threads are used, on a plain weave, there 
would be practically no contraction, as the 
filling is not strong enough to bend them out 
of a straight line. 

On leno and lappet weaves, the rate of con- 
traction is considerably more than the ground 
ends. 

The rate per cent of contraction on these 
cords or ply ends may be found by taking a 
piece of cloth to be reproduced, and, by cut- 
ting a piece of the cloth 3 or 4 inches long, 
pulling the threads out, and measuring the 
length of the thread. 

If a 3 inch piece is taken and the cords 
when pulled out of the cloth measured 4 in- 
ches, the per cent of contraction may be found 
as follows : 



47 

4 inches of yarn wove 3 inches of cloth, 
4 — 3 =: 1 inch for contraction. 
1 X 100 ^ 3 = 33.33 per cent of con- 
traction. 

To find filling required to govern yards 
per lb. when more than 1 size of warp 
yarn is used- 
Rule. Multiply the number of e?ids of each 
warp counts by its own rate of contraction^ divide 
each by its standard yards (for cotton 840) and 
its own counts^ add size to the single yarns ; add 
all the results^ and answer is total weight of warp 
yarns. 

Multiply width at reed by picks per inch, 
and result by length of cloth; divide by 
standard yards (for cotton 840), and answer 
is hanks of filling yarn. 

Divide the length of cloth by yards per lb. ; 
answer is weight of cloth. 

Deduct weight of warp yarns from total 
weight of cloth, and balance is weight for 
filling. 

Divide hanks of filling by weight for filling ; 
answer is counts of filling required. 

When calculating weights, the better way is 
to take a 50 yard length, and base all calcula- 
tions on that length of cloth. 



48 

To find the production of a loom in yards 
per week- 
Rule. Multiply the speed of the loom by one- 
sixth of the hours run per week^ add one cipher 
to the restilt^ and divide this result by the picks 
per inch in the cloth. Answer is yards per week^ 
with no allowance for stoppage. 

Example. Find the production of a loom 
for h% hours, which runs 180 picks per minute 
on a cloth with 64 picks per inch. 

58 -^ 6 =: 9| 

180 X 9| = 1740, add 1 cipher = 17400 

17400 -i- 64 = 271.87 yards per week. 

To find the production of a loom with a 
certain per cent off for stoppage- 
Rule. Multiply the speed of the loojn by one- 
sixth of the hours per week. Multiply the result 
by the per cent of production required., cut off one 
figure at the 7'ight from the result^ and divide by 
picks per inch in the cloth. 

Example. Find the yards per week of 58 
hours a loom will weave, runnino^ 180 picks 
per minute on 72 pick cloth, at 90 per cent 
production. 

58 -^ 6rr= 9f 

180 X 9f X 90% = 156600, less 1 figure 
to the rights 15660. 

15660 -^ 72 = 217.50 yards per week. 



49 

To find the cuts per week, divide the 
yards per week by length of cut. 

To find the speed of a loom- 
Rule. Multiply the speed of driving shaft by 

the size of driving pulley and divide by the size of 

loom pulley. 

Example. What is the speed of a loom 
that has a 13 inch pulley driven by a 8 inch 
puUev on driving shaft that makes 240 R. P. 
M. 

240 X 8 = 1920 

1920 ^ 13 = 147.69 speed of loom. 

To find the size of driving pulley re- 
quired- 
Rule. Multiply the size of loom pulley by the 
speed of loom required., and divide by the revolu- 
tio7is per minnte of driving shaft. 

Example. A loom has a 12^ inch pulley, 
and main shaft runs 250 revolutions per min- 
ute. What size of driving pulley will be re- 
quired to drive the loom 180 picks per minute ? 

12.50 X 180 = 2250 

2250 -^ 250 = 9 inches pulley. 






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T N D EX 



To find the counts of a Cotton Yarn, when the 

length and weight are known, ... 6 

To find the dividend for any length of Yarn, . 8 

To find the length in Yards when the counts and 

weight are known, ..... 9 

To find the weight when the length and counts 

are known, ....... 11 

To find the turns of twist per inch, . . 12 

To find the standard hreaking weight of Warp 

Yarn, 13 

To find the counts when threads of different 

counts are twisted together, .... 14 

To find the counts of a ply thread, when three or 
more threads of different counts are twisted 
together, ....... 14 

To find the counts of a Yarn which must be 
twisted with another to make a Ply Yarn of 
a given count, . . . .15 

To find the weight of each counts of Yarn when 
twisting two threads together to obtain a given 
weight of Ply Yarn, ..... 15 

To find the cost per lb. of a Ply Yarn made from 

two or more single yarns, . . . .16 

To find the number of Dram Silk from a given 

length and weight, ..... 19 

To find the number of yards when the weight and 

counts are known, ...... 19 



52 

To find the dividends for short lengths of worsted 

yarns, 20 

To find counts, length, or weight, apply rules 
given for Cotton Yarns, and use the standard 
for worsted, i. e. 560 j-ards, .... 21 

To find counts, length, or weight, apply rules given 
for Cotton and Worsted Yarns, and use the 
standard length of the system desired, . 22 

To find the equivalent counts of a yarn in any 

system, ........ 23 

To find the price of a Ply Yarn when the threads 
twisted together are of different values, and 
different counts, ...... 24 

To find the number of ends required in a warp 

when the sley and width are given, . . 28 

To find the dents per inch in a reed required for 

any sley cloth, ...... 29 

To find the sley of the cloth a reed will weave 

when length and total dents in reed are known, 30 

To find width of the warp ends in the reed, . 31 

To find the slasher length of a cut for a given 

length of cloth, ...... 32 

To find the weight of Warp Yarn in a piece of 

cloth, 32 

To find the weight of filling, .... 33 

To find average number when sley, pick, width 

and yards per lb. are known, . . . 35 

To find average number when sley, pick, width 

and yards per lb. are known, ... 36 

To find average number when sley, pick, warp 

and filling are knoAvn, ..... 37 



53 

To find average number of warp, when two or 

more different counts are used, ... 37 

To find jards per lb. when sley, pick, counts of 

warp, filling and width are known, . . 38 

To find the yards per lb. from a small piece of 

cloth, . 39 

To find filling required to govern yards per lb., 40 

To find filling required to govern yards per lb., 
when sley, picks, width and yards per lb. are 
known, ........ 40 

To find the number of ends in a warp for a fancy 

cloth of irregular reed draft, ... 41 

To find the number of patterns when the sley 

reed and width of the cloth are known, . 42 

To find the average sley on irregular drafted 

cloths, ........ 45 

To find the sley reed to use to make a cloth any 

average sley desired, ..... 45 

To find filling required to govern yards per lb. 

when more than 1 size of warp yarn is used, 47 

To find the production of a loom in yards per 

week, ........ 48 

To find the production of a loom with a certain 

per cent off for stoppage, .... 48 

To find the cuts per week, divide the yards per 

week by length of cut, ..... 49 

To find the speed of a loom, .... 49 

To find the size of driving pulley required, . 49 

Reed Table, 50 



237 90 



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HECKMAN 

BINDERY INC. 

# MAY 90 
N. MANCHESTER, 
INDIANA 46962 







